为什么没有将正确的 ID 插入到该表中？

Question

我希望下面的代码在customer_payment 表中插入汽车的ID，但它只选择477 作为ID。我不知道为什么。如下图所示，只插入了product_id = 477，没有插入任何其他值。如果我选择 500，它仍然会插入 477。

include 'admin/db.php';

if(isset($_GET['payment_here'])){

    //select product id from cart
    $select_cart = "select * from cart";
    $runcart = mysqli_query($conn, $select_cart);
    $cartwhile=mysqli_fetch_assoc($runcart);

    $carssid = $cartwhile['P_ID'];
    $cusid = $cartwhile['C_ID'];

    //select id from cars
    $scars = "select * from cars where id=$carssid";
    $scarsrun = mysqli_query($conn, $scars);
    $showcars = mysqli_fetch_assoc($scarsrun);
    $carsdealer = $showcars['dealer'];


    //select customer id from customer  table
    //$selectcust = "select * from customer_register where id=$cusid";
    //insert data into customer payment table
    echo $insertpay = "insert into customer_payment 
    (Product_id, customer_id, dealer) 
    values ( $carssid," . $_SESSION['customer_id'] . ", '$carsdealer')";
    $run_inserts = mysqli_query($conn, $insertpay);
    /*
    if($run_inserts){
        echo "<script>window.location.href = 'checkout.php'</script>";
    }
    */
}
?>

为什么没有将正确的 ID 插入到该表中？

Answer 1

你想在这里做什么

$select_cart = "select * from cart";
$runcart = mysqli_query($conn, $select_cart);
$cartwhile=mysqli_fetch_assoc($runcart); // here

正在从“购物车”表中获取第一个条目，该条目始终将是相同的。

你可以试试这样的。

$c_id = $_SESSION['customer_id'];
$select_cart = "select * from cart where C_ID=$c_id";
$runcart = mysqli_query($conn, $select_cart);
$cartwhile=mysqli_fetch_assoc($runcart);

此查询将专门为当前会话的客户获取数据。 其余代码您可以按原样使用。